That sounds a lot like the q-deformed derivative f(qx) - f(x) ------------. qx - x When applied to a monomial f(x) = a x^n, you get a q^n x^n - a x^n -----------------, qx - x which reduces to (1 + q + ... + q^{n-1}) a x^{n-1}. By denoting the sum of powers of q as [n], we get that [n] -> n as q -> 1. But there are lots of interesting things to do with q-derivatives when q is not 1; in particular, when q=p^k, it talks about n-dimensional vector spaces over F_q. John Baez lists the following analogies here: http://math.ucr.edu/home/baez/week259.html q = 1 q = a power of a prime number n-element set (n-1)-dimensional projective space over Fq integer n q-integer [n] permutation groups Sn projective special linear group PSL(n,Fq) factorial n! q-factorial [n]! On Mon, Nov 21, 2011 at 11:58 AM, Henry Baker <hbaker1@pipeline.com> wrote:
We all know that f'(x)=df(x)/dx is the limit of
[f(x+deltax)-f(x)]/deltax
as deltax approaches zero.
However, what if I'm interested in a situation where deltax>0 doesn't approach zero.
Suppose I have a variable y=f(x) and I want to consider a _fixed_ deltay.
Is there any straightforward way to find an x such that deltax is minimized, yet
deltay = f(x+deltax)-f(x)
???
There may be no such pair (x,deltax), or there may be a multiplicity of x's with the same minimal deltax.
I'm primarily interested in finding one, but finding them all would be even better.
Wouldn't you just look for maxima of f'(x)? -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com